On one-sided Lie nilpotent ideals of associative rings

ON ONE-SIDED LIE NILPOTENT IDEALS OF ASSOCIATIVE RINGS

arXiv:0803.0968v1 [math.RA] 6 Mar 2008

VICTORIYA S. LUCHKO AND ANATOLY P. PETRAVCHUK Abstract. We prove that a Lie nilpotent one-sided ideal of an associative ring R is contained in a Lie solvable two-sided ideal of R. An estimation of derived length of such Lie solvable ideal is obtained depending on the class of Lie nilpotency of the Lie nilpotent one-sided ideal of R. One-sided Lie nilpotent ideals contained in ideals generated by commutators of the form [. . . [[r1 , r2 ], . . .], rn−1 ], rn ] are also studied.

Introduction It is well-known that if I is an one-sided nilpotent ideal of an associative ring R then I is contained in a two-sided nilpotent ideal of R. Hence the following question is of interest: for which one-sided ideal I of the ring R there exists a two-sided ideal J such that J ⊇ I and J has properties like properties of I. In [5] it was noted that for an one-sided commutative ideal I of a ring R there exists a nilpotent-by-commutative two-sided ideal J of the ring R such that J ⊇ I. Note that Lie nilpotent and Lie solvable associative rings were investigated by many authors (see, for example [4], [6], [7], [1]) and the structure of such rings is studied well enough. In this paper we prove that a Lie nilpotent one-sided ideal I of an associative ring R is contained in a Lie solvable two-sided ideal J of R. An estimation (rather rough) of Lie derived length of the ideal J depending on Lie nilpotency class of I is also obtained (Theorem 1). In case when the Lie nilpotent one-sided ideal I is contained in the ideal Rn of the ring R generated by all commutators of the form [. . . [[r1 , r2 ], . . .], rn−1 ], rn ] and the Lie derived length of I is less then n it is proved that I is contained in a nilpotent two-sided ideal of R (Theorem 2). The notations in the paper are standard. If S is a subset of an associative ring R then by AnnlR (S) (AnnrR (S)) we denote the left (respectively right) annihilator of S in R. We also denote by R(−) the adjoint Lie ring of (−) the associative ring R. Further, by Rn we denote the n-th member of (−) (−) the lower central series of the Lie ring R(−) . Then Rn = Rn + Rn · R = (−) (−) = Rn +R·Rn is a two-sided ideal of the (associative) ring R. In particular, R2 is a two-sided ideal of the ring R generated by all commutators of the form [r1 , r2 ] = r1 r2 −r2 r1 , r1 , r2 ∈ R. If R is a Lie solvable ring (i.e. such that Date: March 7, 2008. 2000 Mathematics Subject Classification. 16D70. Key words and phrases. associative ring, one-sided ideal, Lie nilpotent ideal, derived length . 1

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R(−) is a solvable Lie ring) then we denote by s(R) its Lie derived length. Analogously, by c(R) we denote Lie nilpotency class of a Lie nilpotent ring R. 1. Lie nilpotent one-sided ideals Lemma 1. Let I be an one-sided ideal of an associative ring R and Z = Z(I) be the center of I. Then there exists an ideal J in R such that J 2 = 0 and [Z, R] ⊆ J. Proof. Let, for example, I be a right ideal from R. Take arbitrary elements z ∈ Z, i ∈ I, r ∈ R. Then it holds z(ir) − (ir)z = 0 (since ir ∈ I). This implies the equality i(zr − rz) = 0 since z ∈ Z(I). As elements z, i, r are arbitrarily chosen then we have I[Z, R] = 0. Consider the right annihilator T = AnnrR (I). It is clear that T is a two-sided ideal of the ring R (since I is a right ideal of R) what implies that [Z, R] ⊆ T . Further, for any element of the form zr − rz from [Z, R] and for any t ∈ T it holds (zr − rz)t = z(rt) − r(zt). Since rt ∈ T then z(rt) = 0. Besides, z ∈ I and therefore zt = 0 what brings the equality (zr − rz)t = 0. It means that [Z, R] · T = 0. Consider the left annihilator J = AnnlT (T ). It is easy to see that J is a two-sided ideal of the ring R. From relations [Z, R] ⊆ T and [Z, R] · T = 0 we have the inclusion [Z, R] ⊆ J. It is also clear that J 2 = 0. Analogously one can consider the case when I is a left ideal. Theorem 1. Let R be an associative ring and I be an one-sided ideal of R. If the subring I is Lie nilpotent then I is contained in a Lie solvable two-sided ideal J of R such that s(J) ⊆ m(m + 1)/2 + m where m = c(I) is Lie nilpotency class of the subring I. Proof. Let for example I be a right ideal. We prove our proposition by the induction on the class of Lie nilpotency n = c(I) of the subring I. If n = 1 then I is a commutative right ideal and by Lemma 1 the ring R contains such an ideal T with zero square that it holds (I + T )/T ⊆ Z(R/T ) in the quotient ring R/T where Z(R/T ) is the center of R/T . It means that I + T is a two-sided ideal of the ring R and s(I + T ) 6 2. Clearly 2 = n + n(n + 1)/2 if n = 1 and the statement of Theorem is true in case n = 1. Assume that the statement is true in case c(I) 6 n − 1 and prove it when c(I) = n. Denote by Z the center of the subring I. By Lemma 1 there exists an ideal T of R with T 2 = 0 such that [Z, R] ⊆ T . Consider the quotient ring R = R/T . Then Z = (Z + T )/T lies in the center of R and therefore Z + Z · R = Z + R · Z is a two-sided ideal of the ring R. Since Z ⊆ I = (I + T )/T the ideal Z + Z · R is Lie nilpotent of and its class of Lie nilpotency 6 m. Further, the quotient ring R/(Z + Z · R) contains the right Lie nilpotent ideal I + (Z + Z · R)/(Z + Z · R) which is Lie nilpotent of class of Lie nilpotency 6 m − 1. By the induction assumption the last right ideal is contained in some Lie solvable ideal of the ring R/(Z + Z · R) of derived length 6 (m−1)m + (m − 1). Since Z + Z · R is Lie solvable and its derived 2 length 6 m (even 6 [log2 m] + 1 but we take a rough estimation) and we consider the quotient ring R/T where T is Lie solvable of derived length 1,


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one can easily see that I is contained in some Lie solvable (two-sided) ideal of derived length which does not exceed (m − 1)m (m + 1)m + (m − 1) + (m + 1) = + m. 2 2 Analogously one can consider the case when I is right ideal. It seems to be unknown whether a sum of two Lie nilpotent associative rings is Lie solvable. So the next statement can be of interest (see also results about sums of P I-rings in [3]). Corollary 1. Let R be an associative ring which can decomposed into a sum R = A + B of its Lie nilpotent subrings A and B. If at least one of these subrings is an one-sided ideal of R then the ring R is Lie solvable. Remark 1. The statements of Theorem 1 and its Corollary become false when we replace Lie nilpotency of one-sided ideals by Lie solvability. Really, consider full matrix ring R = M2 (K) over an arbitrary field K of characteristic 6= 2. It is clear that x y I= x, y ∈ K 0 0

is a right Lie solvable ideal of the ring R but I is not contained in any Lie solvable ideal of R since R is a non-solvable Lie ring. It is also clear that 0 0 R = I + J where J = z, t ∈ K , z t i.e. the simple associative ring R is a sum of two right Lie solvable ideals. 2. On embedding of Lie nilpotent ideals in rings

Lemma 2. Let R be an associative ring, A be a Lie nilpotent subring of R of Lie nilpotency class < m. If Z0 is a subring of A such that Z0 ⊆ Z(R) and Z0 R ⊆ A then Z0m Rm = 0. Proof. Consider the two-sided ideal J = Z0 + Z0 R = Z0 + RZ0 of the ring R. As J ⊆ A then [J, ..., J] = 0 by the condition c(A) 6 m. Further, it is | {z } m

easily to show that

[J, J] = [Z0 + Z0 R, Z0 + Z0 R] = Z02 [R, R]. By induction on k one can also show that [J, ..., J] = Z0k [R, ..., R]. Then | {z } | {z } k

k

we have from the condition on J that [J, ..., J] = Z0m [R, ..., R] = 0. This | {z } | {z } m

m

implies the equality

Z0m Rm = Z0m ([R, ..., R] + [R, ..., R] ·R) = [J, ..., J] + [J, ..., J] ·R = 0. | {z } | {z } | {z } | {z } m

m

m

m

Lemma 3. Let R be an associative ring, I be an ideal of R. Then 1) if J is a nilpotent ideal of the subring I then J lies in a nilpotent ideal JI of the ring R such that JI ⊆ I;

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2) if S = AnnlI (I) (or AnnrI (I)) then S is contained in a nilpotent ideal of the ring R which is contained in I. The proof of this Lemma immediately follows from Lemma 1.1.5 from [2]. Theorem 2. Let R be an associative ring and I be a Lie nilpotent one-sided ideal of R. If I ⊆ Rn and Lie nilpotency class of I is less then n then I is contained in an (associative) nilpotent ideal of R. Proof. Let for example I be a right ideal of the ring R and I ⊆ Rn . One can assume that that n > 2 because the statement of Theorem is obvious in case n = 1. We fix n > 2 and prove the statement of Theorem by induction on the class of Lie nilpotency c = c(I) of the subring I. If c = 0 then I is the zero ideal. and the proof is complete. Assume that the statement is true for rings R with c(I) 6 c − 1 and prove it in case c(I) = c. Since I is Lie nilpotent then by Lemma 1 there exists a nilpotent ideal T of the ring R such that in the quotient ring R = R/T it holds [Z0 , R] = 0 where Z0 is the center of the subring I and Z0 = (Z0 + T )/T . Then by Lemma 2 it holds the relation Z0n · Rn = 0. If Z0n = 0 then Z0 + Z0 R is a nilpotent ideal of the ring R and then the subring Z0 is contained in the nilpotent ideal J = Z0 + T of the ring R. Since in the quotient ring R/J for the right ideal (I + J)/J it holds the inequality c((I + J)/J) 6 c − 1 then by the inductive assumption (I + J)/J is contained in a nilpotent ideal S/J of the ring R/J. But then I ⊆ S where S is nilpotent ideal of the ring R. Let now Z0n 6= 0. Then Z0n ⊆ AnnlR (Rn ) and since Z0 ⊆ Rn then Z0n is n

contained in a nilpotent ideal M of the ring R by Lemma 3. It is obvious that Zo is nilpotent ideal of the ring R. Repeating the above considerations we see that I ⊆ S where S is a nilpotent ideal of the ring R. Corollary 2. Let R be an associative ring with condition R = [R, R]. If I is a Lie nilpotent one-sided ideal of R then there exists a nilpotent (two-sided) ideal J of the ring R such that I ⊆ J Corollary 3. Let R be a semiprime ring. Then every Lie nilpotent onesided ideal is contained in the center Z(R) of the ring R and has trivial intersection with the ideal R2 . Proof. Really since all nilpotent ideals of the ring R are zero then by Lemma 1 every Lie nilpotent one-sided ideal I is contained in Z(R). Since IR ⊆ Z then [IR, R] = I[R, R] = 0. Then from this equality we have IR2 = I([R, R] + [R, R] · R) = 0. Denote J = I ∩ R2 . It is easily to show that J ⊆ AnnlR2 (R2 ) and by Lemma 3 the intersection J lies in a nilpotent ideal of the ring R. As the ring R is semiprime we have J = 0. References [1] B.Amberg and Ya.P.Sysak, Associative rings with metabelian adjoint group, Journal of Algebra, 277 (2004), 456-473. [2] V.A.Andrunakievich and Yu.M.Ryabukhin, Radicals of algebras and structure theory, Nauka, Moscow, 1979. (in Russian).


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[3] B.Felzenszwalb, A.Giambruno and G.Leal, On rings which are sums of two P Isubrings: a combinatorial approach, Pacific Journal of Math., 209, no.1 (2003), 1730. [4] S.A.Jennigs, On rings whose associated Lie rings are nilpotent, Bull. Amer. Math. Soc., 53 (1947), 593-597. [5] A.P.Petravchuk, On associative algebras which are sum of two almost commutative subalgebras, Publicationes Mathematicae (Debrecen). 53, no.1-2, (1998), 191-206. [6] R.K.Sharma and I.B.Srivastava, Lie solvable rings, Proc. Amer. Math. Soc., 94, no.1 (1985), 1-8. [7] W.Streb, Ueber Ringe mit aufloesbaren assoziirten Lie-Ringen, Rendiconti del Seminario Matematico dell’Universit` a di Padova, 50, (1973), 127-142. Algebra Department, Faculty of Mechanics and Mathematics, Kyiv Taras Shevchenko University, Volodymyrskaia street, 01033 Kyiv, Ukraine E-mail address: [email protected] Algebra Department, Faculty of Mechanics and Mathematics, Kyiv Taras Shevchenko University, Volodymyrskaia street, 01033 Kyiv, Ukraine E-mail address: [email protected]